Question

10 numeric 10 points 1.50 g of nitrogen monoxide react with 2.00 g of oxygen gas to produce nitrogen trioxide according to equation 1. equation 1. no (g) + o₂ (g) → no₃ (g) after all of the no has reacted, the excess o₂ is then reacted with 1.00 g of br₂ to produce bromine monoxide according to equation 2. equation 2. o₂ (g) + br₂ (l) → 2 bro (l) calculate how many grams of bro will be produced. answer

Explanation:

Step1: Balance equation 1

The balanced equation 1 is $2NO(g)+3O_2(g)
ightarrow 2NO_3(g)$.

Step2: Calculate moles of NO

The molar - mass of $NO$ is $M_{NO}=14.01 + 16.00=30.01\ g/mol$. The moles of $NO$, $n_{NO}=\frac{m_{NO}}{M_{NO}}=\frac{1.50\ g}{30.01\ g/mol}=0.05\ mol$.

Step3: Calculate moles of $O_2$ required for NO reaction

From the balanced equation 1, the mole - ratio of $NO$ to $O_2$ is $2:3$. So the moles of $O_2$ required for the reaction with $NO$, $n_{O_2\ required}=\frac{3}{2}n_{NO}=\frac{3}{2}\times0.05\ mol = 0.075\ mol$.

Step4: Calculate moles of $O_2$ initially

The molar - mass of $O_2$ is $M_{O_2}=2\times16.00 = 32.00\ g/mol$. The moles of $O_2$ initially, $n_{O_2\ initial}=\frac{m_{O_2}}{M_{O_2}}=\frac{2.00\ g}{32.00\ g/mol}=0.0625\ mol$.

Step5: Determine the limiting reactant in equation 2

Since $n_{O_2\ initial}ightarrow 2NO_2(g)$. The moles of $O_2$ required for $0.05\ mol$ of $NO$ is $n_{O_2\ required}=\frac{1}{2}n_{NO}=\frac{1}{2}\times0.05\ mol = 0.025\ mol$. The moles of excess $O_2$, $n_{O_2\ excess}=n_{O_2\ initial}-n_{O_2\ required}=0.0625\ mol - 0.025\ mol=0.0375\ mol$.
The molar - mass of $Br_2$ is $M_{Br_2}=2\times79.90 = 159.80\ g/mol$. The moles of $Br_2$, $n_{Br_2}=\frac{m_{Br_2}}{M_{Br_2}}=\frac{1.00\ g}{159.80\ g/mol}\approx0.00626\ mol$.
From the balanced equation 2 $O_2(g)+Br_2(l)
ightarrow 2BrO(l)$, the mole - ratio of $O_2$ to $Br_2$ is $1:1$. Since $n_{Br_2}<n_{O_2\ excess}$, $Br_2$ is the limiting reactant.

Step6: Calculate moles of BrO

From the balanced equation 2, the mole - ratio of $Br_2$ to $BrO$ is $1:2$. So the moles of $BrO$, $n_{BrO}=2n_{Br_2}=2\times0.00626\ mol = 0.01252\ mol$.

Step7: Calculate mass of BrO

The molar - mass of $BrO$ is $M_{BrO}=79.90 + 16.00=95.90\ g/mol$. The mass of $BrO$, $m_{BrO}=n_{BrO}\times M_{BrO}=0.01252\ mol\times95.90\ g/mol\approx1.19\ g$.

Answer:

$1.19\ g$